Let $S$ be a surface in 3D described by the equation: $2y - 3z + \ln(x + y) - \dfrac{3}{2} = \ln\left( \dfrac{1}{4} \right)$ Fill in the rest of the equation of the plane tangent to $S$ at $\left( \dfrac{-1}{2}, \dfrac{3}{4}, 0 \right)$.
Answer: The equation for a tangent plane of an implicitly defined surface $F(x, y, z) = 0$ at the point $(a, b, c)$ is: $F_x(x - a) + F_y(y - b) + F_z(z - c) = 0$ [What's the intuition behind the formula?] We can see from the formula that the three values we're missing are $F_x$, $F_y$, and the value to subtract from the $z$ -coordinate. The formula should make the $z$ -term zero at $z = 0$, so we want to subtract $0$ from $z$. $\begin{aligned} F_x &= \dfrac{1}{x + y} = \dfrac{1}{\dfrac{-1}{2} + \dfrac{3}{4}} = 4 \\ \\ F_y &= 2 + \dfrac{1}{x + y} = 2 + 4 = 6 \end{aligned}$ Here's the completed equation for the tangent plane of $S$ at $\left( \dfrac{-1}{2}, \dfrac{3}{4}, 0 \right)$ : $4 \left( x + \dfrac{1}{2} \right) + 6 \left( y - \dfrac{3}{4} \right) - 3 (z - 0) = 0$